Giả sử có 100g đá

=> \(m_{CaCO_3}=\dfrac{100.80}{100}=80\left(g\right)\)

\(n_{CaCO_3}=\dfrac{80}{100}=0,8\left(mol\right)\)

Gọi số mol CaCO₃ phân hủy

PTHH: CaCO₃ --t^o--> CaO + CO₂

             a-------------->a--->a

=> m_Y = 100 - 44a (g)

=> m_CaO = 56a (g)

=> \(\dfrac{56a}{100-44a}.100\%=45,65\%\)

=> a = 0,6 (mol)

=> \(H=\dfrac{0,6}{0,8}.100\%=75\%\)

\(1)PTHH:CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow\\ n_{CaCO_3}=\dfrac{500.95\%}{100}=4,75(mol)\\ \Rightarrow n_{CaO}=4,75(mol)\\ \Rightarrow m_{CaO}=4,75.56=266(g)\\ \Rightarrow m_{CaO(tt)}=266.80\%=212,8(g)\\ m_{CaCO_3(k p/ứ)}=500.95\%.20\%=95(g)\\ \Rightarrow m_A=95+212,8=307,8(g)\\ 2)\%m_{CaO}=\dfrac{212,8}{307,8}.100\%=69,136\%\\ n_{CO_2}=n_{CaO}=4,75(mol)\\ \Rightarrow V_{CO_2}=4,75.22,4=106,4(l)\)

\(m_{CaCO_3}=90\%.400=360\left(g\right)\\ \rightarrow n_{CaCO_3}=\dfrac{360}{100}=3,6\left(mol\right)\)

PTHH: CaCO₃ --t^o--> CaO + CO₂

             3,6 ----------> 3,6 -----> 3,6

\(\rightarrow n_{CaO}=3,6.75\%=2,7\left(mol\right)\\ \rightarrow n_{CaCO_3\left(chưa.pư\right)}=3,6-2,7=0,9\left(mol\right)\)

\(\rightarrow m_X=0,9.100+2,7.56=241,2\left(g\right)\\ \%m_{CaO}=\dfrac{0,9.100}{241,2}=37,31\%\)

\(V_Y=V_{CO_2}=3,6.75\%.22,4=60,48\left(l\right)\)

\(m_{CaCO_3}=\dfrac{400\cdot90\%}{100\%}=360g\Rightarrow n_{CaCO_3}=\dfrac{360}{100}=3,6mol\)

\(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

3,6           3,6       3,6

Thực tế: \(n_{CaO}=3,6\cdot75\%=2,7mol\)

\(\Rightarrow m_{CaO}=2,7\cdot56=151,2g\)

a) \(n_{CaCO_3\left(bđ\right)}=\dfrac{400.90\%}{100}=3,6\left(mol\right)\)

=> \(n_{CaCO_3\left(pư\right)}=\dfrac{3,6.75}{100}=2,7\left(mol\right)\)

PTHH: CaCO₃ --t^o--> CaO + CO₂

              2,7---------->2,7---->2,7

=> m_X = m_{đá vôi} - m_CO2 = 400 - 2,7.44 = 281,2 (g)

b) 

m_CaO = 2,7.56 = 151,2 (g)

=> \(\%CaO=\dfrac{151,2}{281,2}.100\%=53,77\%\)

V_CO2 = 2,7.22,4 = 60,48 (l)

1)

1,2 tấn = 1200(kg)

5 tạ = 500(kg)

\(m_{CaCO_3} = 1200.80\% = 960(kg)\)

\(CaCO_3 \xrightarrow{t^o} CaO + CO_2\\ n_{CaCO_3\ pư} = n_{CaO} = \dfrac{500}{56}(mol)\\ \Rightarrow H = \dfrac{\dfrac{500}{56}.100}{960}.100\% = 93\%\)

1 (H)= 93,11%

2 (H)=88.08%

m cao=1.064(tấn)

==> m cr = 1.065(tấn)

%m cao = 56%

mCaCO3= 80%. m(đá vôi)= 80%. 500=400(g)

-> nCaCO3= mCaCO3/M(CaCO3)=400/100=4(mol)

PTHH: CaCO3 -to-> CaO + CO2

Ta có: nCaO(LT)= nCaCO3= 4(mol)

=> mCaO(LT)=56.4=224(g)

Đặt x là số mol CaCO3 (p.ứ) -> Số mol CaO tạo thành là x (mol) (x>0)

=> Khối lượng rắn tạo thành là: 

(400 - 100x) + 56x + 100= 78%. 500

<=>x=2,5(mol) 

Vì KL tỉ lệ thuận số mol:

=> H(p.ứ)= (2,5/4).100= 62,5%

Câu 7:

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Vì:\dfrac{0,4}{1}>\dfrac{0,3}{1}\Rightarrow H_2SO_4dư,Znhết\\ a,n_{H_2}=n_{Zn}=0,3\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,n_{O\left(mất\right)}=n_{H_2O}=n_{H_2}=0,3\left(mol\right)\\ \Rightarrow m_{giảm}=m_{O\left(mất\right)}=0,3.16=4,8\left(g\right)\\ \Rightarrow m=4,8\left(g\right)\)

Câu 6:

- Giả sử có 1 mol hỗn hợp khí A. 

\(\Rightarrow\left\{{}\begin{matrix}n_{N_xO}=30\%.1=0,3\left(mol\right)\\n_{SO_2}=30\%.1=0,3\left(mol\right)\\n_{CO_2}=1-\left(0,3+0,3\right)=0,4\left(mol\right)\end{matrix}\right.\\ \Rightarrow m_A=0,3.\left(14x+16\right)+0,3.64+0,4.44=41,6+4,2x\left(g\right)\\ \%m_{N_xO}=19,651\%\\ \Leftrightarrow\dfrac{4,2x+4,8}{41,6+4,2x}.100\%=19,651\%\\ \Leftrightarrow x=1\\ \Rightarrow N_xO.là:NO\\ M_{hhA}=\dfrac{0,3.30+0,3.64+0,4.44}{1}=45,8\left(\dfrac{g}{mol}\right)\\ \Rightarrow d_{\dfrac{hhA}{H_2}}=\dfrac{45,8}{2}=22,9\)

Câu 4:

\(a,m_{CaCO_3\left(ban.đầu\right)}=95\%.m_{đá.vôi}=95\%.500=475\left(g\right)\\ \Rightarrow n_{CaCO_3\left(ban.đầu\right)}=\dfrac{475}{100}=4,75\left(mol\right)\\ PTHH:CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\\ n_{CaO\left(LT\right)}=n_{CaCO_3\left(ban.đầu\right)}=4,75\left(mol\right)\\ Vì:H=80\%\Rightarrow n_{CaO\left(TT\right)}=4,75.80\%=3,8\left(mol\right)\\ \Rightarrow m_{rắnA}=m_{tạp.chất}+m_{CaCO_3\left(không.p.ứ\right)}+m_{CaO\left(TT\right)}\\ =25+\left(4,75-3,8\right).100+3,8.56=332,8\left(g\right)\\ b,\%m_{CaO}=\dfrac{3,8.56}{332,8}.100\approx63,942\%\\ n_B=n_{CO_2\left(TT\right)}=n_{CaO\left(TT\right)}=3,8\left(mol\right)\\ \Rightarrow V_{B\left(đktc\right)}=V_{CO_2\left(đktc\right)}=3,8.22,4=85,12\left(l\right)\)

\(m_{CaCO_3}=400\cdot90\%=360\left(g\right)\)

\(m_{trơ}=400-360=40\left(g\right)\)

\(n_{CaCO_3}=\dfrac{360}{100}=3.6\left(mol\right)\)

\(a.\)

\(n_{CaCO_3\left(pư\right)}=3.6\cdot75\%=2.7\left(mol\right)\)

\(CaCO_3\underrightarrow{^{^{t^0}}}CaO+CO_2\)

\(2.7........2.7...........2.7\)

\(m_X=m_{CaO}+m_{CaCO_3\left(dư\right)}+m_{trơ}=2.7\cdot56+\left(3.6-2.7\right)\cdot100+40=281.2\left(g\right)\)

\(b.\)

\(\%CaO=\dfrac{2.7\cdot56}{281.2}\cdot100\%=53.77\%\)

\(V_{CO_2}=2.7\cdot22.4=60.48\left(l\right)\)

Chọn đáp án C 

m C a C O 3 = 0 , 8 m

C a C O 3 ® CaO + C O 2

    a             a         a   (mol) 

44a = m – 0,78m Þ a = 0,005m

H % = 100 . 0 , 005 m . 100 % 0 , 8 m = 62 , 5 %